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Sergeeva-Olga [200]
1 year ago
14

Based on 20 samples, the average resistance is 25 ohms and the sample standard deviation is 0.5 ohm. Determine the 90% confidenc

e interval of the standard deviation of the batch. (10 points)
Mathematics
1 answer:
suter [353]1 year ago
3 0

Answer:

90% Confidence Interval = [24.81, 25.19]

Step-by-step explanation:

From the question, our number of samples is 20, it is small because it is less than 30. Hence we use the t score confidence interval formula

= Mean ± t score × Standard deviation/√n

Mean = 25 ohms

Standard deviation = 0.5 ohms

n = 20

We find the degrees of freedom = n - 1

= 20 - 1 = 19

Using the T score table

T score for 90% confidence interval with degrees of freedom 19

= 1.729

Hence:

Confidence Interval =

= 25 ± 1.729 × 0.5/√20

= 25 ± 0.1933080767

=

Confidence Interval

25 - 0.1933080767

= 24.806691923

≈ 24.81

25 + 0.1933080767

= 25.1933080767

≈ 25.19

90% Confidence Interval = [24.81, 25.19]

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svetoff [14.1K]
Start by finding the sum of (-8 3/4) and (-2 5/6).  Find the least common denominator (LCD).  12 is the first thing that both 4 and 6 will evenly divide into, so we convert both fractions to 12ths:

-8 3/4 = -8 9/12 (multiply 4 by 3 to get 12, so multiply the numerator, 3, by 3 as well)
-2 5/6 = -2 10/12 (multiply 6 by 2 to get 12, so multiply the numerator, 5, by 2 as well)
-8 9/12 + -2 10/12 = -10 19/12
We must convert the improper fraction.  12 goes into 19 1 time with 7 left over, so 19/12 = 1 7/12
This means that -10 19/12 = -11 7/12
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4 0
2 years ago
A 2-column table with 6 rows. The first column is labeled miles driven with entries 27, 65, 83, 109, 142, 175. The second column
Helen [10]

Answer:

1) The linear regression model is y = -0.0348·x + 13.989

2) The correlation coefficient is -0.0725

3) The strength of the model is strong - association

Step-by-step explanation:

1)

                         X            Y          XY       X²

                         27            13           351          729

                         65             12          780         4225

                         83              11          913         6889

                         109            10          1090      11881

                         142            9            1278     20164

                         175              8           1400      30625

                ∑      601              63 5812 74513

From y = ax + b, we have

a = \frac{n\sum xy - \sum x\sum y }{n\sum x^{2}-\left (\sum x  \right )^{2}} = \frac{6 \times 5812  - 601 \times 63}{6 \times 74513-601^{2}} = - 0.0348

b = 1/n(∑y -a∑x) = 1/6(63 - (0.0348)×601) = 13.989

Therefore, the linear regression model is y = -0.0348·x + 13.989

2)

r = \frac{n\sum xy - \sum x\sum y }{\sqrt{[n\sum x^{2}-\left (\sum x  \right )^{2}] [n\sum y^{2}-\left (\sum y  \right )^{2}]}}  = \frac{6 \times 5812  - 601 \times 63}{\sqrt{[6 \times 74513-601^{2}] [6  \times 3969 - 63^2]} } = - 0.0725

3) The strength is - association.

5 0
2 years ago
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Answer:

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Step-by-step explanation:

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Answer:

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